InterviewSolution
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InterviewSolution
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Let `P(a sectheta, btantheta) and Q(aseccphi , btanphi)` (where `theta+phi=pi/2` be two points on the hyperbola `x^2/a^2-y^2/b^2=1` If `(h, k)` is the point of intersection of the normals at `P and Q` then `k` is equal to (A) `(a^2+b^2)/a` (B) `-((a^2+b^2)/a)` (C) `(a^2+b^2)/b` (D) `-((a^2+b^2)/b)`A. `(a^(2)+b^(2))/(a)`B. `-((a^(2)+b^(2))/(a))`C. `(a^(2)+b^(2))/(b)`D. `-((a^(2)+b^(2))/(b))` |
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Answer» Firstly, we obtain the slope fo normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 " at " (a sec theta , b tan theta)`. On differentiating w.r.t.x, we get `(2x)/(a^(2))-(2y)/(b^(2))xx(dy)/(dx)=0 rArr(dy)/(dx)=(b^(2))/(a^(2))(x)/(y)` Slope for normal at the point `(a sec theta , b tan theta )` will be `-(a^(2)b tan theta )/(b^(2)a sec theta)= -(a)/(b) sin theta ` ` therefore ` Equation of normal at `(a sec theta, b tan theta )` is `y-b tan theta= -(a)/(b) sin theta (x-a sec theta)` `rArr (a sin theta )x + "by"=(a^(2)+b^(2))tan theta` `rArr ax+b "cosec" theta =(a^(2)+b^(2)) sec theta " ...(i)" ` Similarly, equation of normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` at `(a sec phi, b tan phi) " is " ax+b y " cosec "phi=(a^(2)+b^(2))sec phi " ... (ii)" ` On subtracting Eq. (ii) from Eq. (i), we get `b("cosec" theta -"cosec" phi)y=(a^(2)+b^(2))(sec theta-sec phi)` `rArr y=(a^(2)+b^(2))/(b)*(sec theta-sec phi)/("cosec"theta-"cosec"phi)` But `(sec theta-sec phi)/("cosec"theta-"cosec"phi)=(sec theta-sec(pi//2-theta))/("cosec"theta-"cosec"(pi//2-theta)) " " [ because phi+theta=pi//2]` `=(sec theta -"cosec"theta)/(sec theta-sec theta)= -1` Thus, `y= -((a^(2)+b^(2))/(b)),i.e.k=-((a^(2)+b^(2))/(b))` |
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